Unveiling the Mystery of Simultaneity

[Santural]

Simultaneity

Let's imagine there is a train moving at the speed of light. When in the middle, two lightning flashes appear on it's left and right. Why is it that it sees the right flash before the left?
And if it is moving left to right, why does it see the left before the right?

It is to do with the train's motion? It's position relative to the flash? If so, how?

P.S: This is not a homework question, just a general question.

 

[bernhard.rothenstein]

train at the speed of light?

Let's imagine there is a train moving at the speed of light. When in the middle, two lightning flashes appear on it's left and right. Why is it that it sees the right flash before the left?
And if it is moving left to right, why does it see the left before the right?

It is to do with the train's motion? It's position relative to the flash? If so, how?

P.S: This is not a homework question, just a general question.

We should have a hard imagination in order to accept that a train could move at the speed of light

 

[pervect]

As has been mentioned once or twice or even more times before, physical objects can't move at the speed of light.

Attempting to sneak the idea that they can in through an assumption in a question leads to confusion and long pointless threads. Technically, it's an example of a fallacy, known as the "loaded question", see for instance

http://en.wikipedia.org/wiki/Fallacy_of_many_questions

What may be confusing to the Original Poster is why objects can't move at the speed of light, or even faster. This is a much more productive issue to discsus, my favorite explanation here involves the way velocities add. But I won't go into much more detail unless someone indicats an interest, and can't find it in one of the other threads where it has been talked about.

 

[Santural]

Well, in that case, let's imagine the train is moving at 0.8886c.

 

[neutrino]

What do you mean by "when in the middle"? Who is the observer here and how is that person moving relative to the train?

 

[Santural]

Slightly changed it for better understanding.
A rocket ship traveling left to right and near the speed of light relative to the Earth observes two lights flashing on Earth (simultaneously). If the rocket observer is directly between them when they flash, the observer sees the one on the right flash before the one on the left.
If the rocket ship in the above question is moving right to left the one on the left flashes before the one on the right.

Why?

 

[matheinste]

If the observer sees the flashes as simultaneous how can he see one before the other.

Matheinste.

 

[JesseM]

Slightly changed it for better understanding.
A rocket ship traveling left to right and near the speed of light relative to the Earth observes two lights flashing on Earth (simultaneously). If the rocket observer is directly between them when they flash, the observer sees the one on the right flash before the one on the left.
If the rocket ship in the above question is moving right to left the one on the left flashes before the one on the right.

Why?

If the flashes were simultaneous in the Earth's frame, they're not simultaneous in the rocket's frame--for example, if the rocket is moving to the right in the Earth's frame, in the rocket's frame the light on the right side flashed at an earlier time than the light on the left side. If the rocket has clocks at the front and back which are synchronized in its own frame using the Einstein synchronization convention (for example, the rocket-observer could have synchronized them by setting off a flash at the center of the rocket, and making sure both clocks read the same time when the light from the flash reaches each one, since light is assumed to travel at c in both directions in the rocket's frame and should therefore take the same time to reach both clocks), then in the Earth-frame these two clocks will be measured to be out-of-sync (in the Earth's frame, if a flash is set off at the center of the rocket and the rocket is moving right, naturally the light will catch up to the left clock before it catches up to the right clock, under the assumption that light travels at c in both directions in the Earth's frame).

 

[JesseM]

If the observer sees the flashes as simultaneous how can he see one before the other.

Matheinste.

If I'm understanding the scenario, "simultaneously" referred to when the flashes happened in the Earth's frame, "sees" referred to when the light from each flash actually reached the rocket-observer's eyes.

 

[pervect]

Well, in that case, let's imagine the train is moving at 0.8886c.

A much better question! I think it has been answered fairly well by other posters.

 

[pervect]

If I'm understanding the scenario, "simultaneously" referred to when the flashes happened in the Earth's frame, "sees" referred to when the light from each flash actually reached the rocket-observer's eyes.

Right, that's what I would have assumed.

Simultaneous events are events that are assigned the same time coordinate, in some particular coordinate system.

Seen is ambiguous, but usually means the time assigned to the arrival of light.

The important point, which can't be stressed enough, is that the set of events regarded as simultaneous is one coordinate system is not simultaneous in another.

This can be drawn on a space-time diagram, for instance the dotted and dashed lines in the diagram below represent the set of events regarded as simultaneous by two different observers. Visual inspection shows that they are not the same set of points.

 

[Santural]

Thanks! I understand, but let me clarify in case of any errors:
-Simultaneity is true only to a certain rest frame.
-In any other rest frame, observers would not experience similar simultaneity as the other frame.

Thanks! And my mistake, the Speed of Light shall be for light alone.

P.S. JesseM, you got the essence of the question, I should have posted it better.

 

[JesseM]

Thanks! I understand, but let me clarify in case of any errors:
-Simultaneity is true only to a certain rest frame.
-In any other rest frame, observers would not experience similar simultaneity as the other frame.

Thanks! And my mistake, the Speed of Light shall be for light alone.

P.S. JesseM, you got the essence of the question, I should have posted it better.

Yup, I think you've got it. To give you a formula, if there are two clocks which are at rest with respect to each other and a distance of x apart in their own frame, and the clocks are synchronized in their own frame, then if in my frame the two clocks are moving at speed v along the axis between them, I will measure the back clock's time to be vx/c^2 ahead of the front clock's time.

 

[Santural]

Gotcha! Thanks for the formula!

 

[paw]

...To give you a formula, if there are two clocks which are at rest with respect to each other and a distance of x apart in their own frame, and the clocks are synchronized in their own frame, then if in my frame the two clocks are moving at speed v along the axis between them, I will measure the back clock's time to be vx/c^2 ahead of the front clock's time.

JesseM, I'm not sure I follow how the position of your co-moving clocks affects the time measured between them in your frame.

Imagine two spaceships (a and b) traveling on a parallel course away from Earth at 0.5c measured in the Earth centered frame. The distance between ships is negligable and perpendicular to the direction of travel. Synchronized clocks are located on the Earth, ship a and ship b.

In the frame where ship a is at rest, Earth is seen to be moving away at 0.5c and a clock on Earth would be seen to be running 25% slow. Ship b is at rest in ship a's frame so it too should observe Earth clock running 25% slow.

Additionally, since ship b is at rest in a's frame it's clock should be observed to be syncronous with ship a's clock. Do we agree so far?

In the Earth centered frame ship a's clock is observed to be running 25% slow. So is ship b's clock. The same argument holds true even if the ships are moving along the axis between then doesn't it? I mean, I don't see any 'x' in the Lorentz transform.

So where does the position (x) come in? I can't see how position can matter as long as ship a and ship b are at rest wrt each other. Am I missing something?

 

[neutrino]

The same argument holds true even if the ships are moving along the axis between then doesn't it?

No, it does not. If the clocks are separated by a distance x (in the clocks' rest frame) along their direction of motion, then the leading clock lags behind trailing clock by an amount xv/c², where v is their relative speed with respect to the observer.

 

[Johnny R]

Slightly changed it for better understanding.
A rocket ship traveling left to right and near the speed of light relative to the Earth observes two lights flashing on Earth (simultaneously). If the rocket observer is directly between them when they flash, the observer sees the one on the right flash before the one on the left.
If the rocket ship in the above question is moving right to left the one on the left flashes before the one on the right.

Why?

Because the speed of light is c and not instantaneous. The same reason that if a person in front of you is running from left to right between two baseball players and they both throw a baseball at the same speed towards the person, the ball thrown from the right will nail the person in the head first.

 

[JesseM]

JesseM, I'm not sure I follow how the position of your co-moving clocks affects the time measured between them in your frame.

Imagine two spaceships (a and b) traveling on a parallel course away from Earth at 0.5c measured in the Earth centered frame. The distance between ships is negligable and perpendicular to the direction of travel. Synchronized clocks are located on the Earth, ship a and ship b.

In the frame where ship a is at rest, Earth is seen to be moving away at 0.5c and a clock on Earth would be seen to be running 25% slow. Ship b is at rest in ship a's frame so it too should observe Earth clock running 25% slow.

Additionally, since ship b is at rest in a's frame it's clock should be observed to be syncronous with ship a's clock. Do we agree so far?

In the Earth centered frame ship a's clock is observed to be running 25% slow. So is ship b's clock. The same argument holds true even if the ships are moving along the axis between then doesn't it?

I think you're confusing "synchronized" with "ticking at the same rate". The Earth does indeed observe each clock to be running at the same rate, but that's not what I'm talking about. Two clocks can be running at the same rate but out-of-sync--for example, one clock can be consistently 1 hour behind the other, so that when the first clock reads 2 the second reads 1, when the first clock reads 3 the second reads 2, and so forth. The clocks of the two ships a and b will be out-of-sync in this sense, assuming they synchronized their clocks in their own frame using the Einstein synchronization procedure. The reason is that the synchronization procedure assumes the speed of light is the same in all directions in the ship's rest frame, so that they could synchronize their clocks by setting off a flash at the midpoint of the line between them, and making sure both clocks read the same time when the light from the flash reaches them. Since the Earth assumes that light moves the same speed in both directions in its frame, this means that in the Earth's frame the light must reach one clock before the other, since in the Earth's frame one ship is moving towards the point the flash happened and one is moving away from that point. This means that if the ships set their clocks to read the same time at the moment the light from the flash hits them, in the Earth's frame one clock's time will naturally be ahead of the other.

I mean, I don't see any 'x' in the Lorentz transform.

Are you sure you're not thinking of the time dilation equation? The Lorentz transform definitely has an x coordinate in it:

[tex]x' = \gamma (x - vt)[/tex]
[tex]y' = y[/tex]
[tex]z' = z[/tex]
[tex]t' = \gamma (t - vx/c^2)[/tex]
where [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

 

[JesseM]

Because the speed of light is c and not instantaneous. The same reason that if a person in front of you is running from left to right between two baseball players and they both throw a baseball at the same speed towards the person, the ball thrown from the right will nail the person in the head first.

That's not a good analogy, because in the frame of the runner, the two balls have different speeds. And if you forget about the relativity of simultaneity, and falsely imagine that the two flashes also happened at the same time in the train's frame, then since both light signals should have a velocity of c in the train's frame, you should predict both light signals reach the center at the same moment.

 

[paw]

No, it does not. If the clocks are separated by a distance x (in the clocks' rest frame) along their direction of motion, then the leading clock lags behind trailing clock by an amount xv/c², where v is their relative speed with respect to the observer.

Ok, I think I see what you're saying. In the Earth (observer) frame, both clocks are moving, so light takes a little less time traveling from a to b than it does from b to a (assuming a is the lead clock). So...

In the Earth (observer) frame, if they are perpendicular to the direction of travel the ships clocks would agree?

At any angle in between they'd disagree by some factor between 0 and xv/c² given by the angle?

This effect must disappear in the limit as x approaches zero, correct?

Now I thought time dilation was completely reciprocal. Since ship a and ship b are at rest wrt each other there is no time dilation effect between them. So how is this reconciled with the fact that an Earth observer does measure a's clock as lagging behind b's?

 

[JesseM]

Ok, I think I see what you're saying. In the Earth (observer) frame, both clocks are moving, so light takes a little less time traveling from a to b than it does from b to a (assuming a is the lead clock). So...

In the Earth (observer) frame, if they are perpendicular to the direction of travel the ships clocks would agree?

At any angle in between they'd disagree by some factor between 0 and xv/c² given by the angle?

This effect must disappear in the limit as x approaches zero, correct?

Yes, all of these are true, assuming the clocks have been synchronized in the ship's frame using the Einstein synchronization convention.

Now I thought time dilation was completely reciprocal. Since ship a and ship b are at rest wrt each other there is no time dilation effect between them. So how is this reconciled with the fact that an Earth observer does measure a's clock as lagging behind b's?

Like I said in my last post, time dilation is only about the rate a clock is ticking. In the Earth's frame, a and b's clock are both ticking at exactly the same rate, it's just that they aren't synchronized. For every hour that a's clock advances, b's clock advances exactly an hour too.

 

[paw]

I think you're confusing "synchronized" with "ticking at the same rate".

Thanks JesseM, you're right. I think I've got it.

Ship a and ship b's clocks run at the same rate wrt to the observer but a lags behind b because it's further from the observer. That makes sense.

The effect would be reciprocal in that from the ships frame a will see the Earth's clock running the same rate as b sees but a will see the Earth clock lagging because it's further from the observer.

Further, the lag between clocks would tend toward zero in the limit as x goes to 0 and as the angle of the ships to the direction of travel goes to 90deg.

It all makes sense now.

And yes I did mean the time dilation eq rather than the Lorentz tranform. It's been a while since I've studied physics so I'm a little fuzzy...

 

[Johnny R]

That's not a good analogy, because in the frame of the runner, the two balls have different speeds. And if you forget about the relativity of simultaneity, and falsely imagine that the two flashes also happened at the same time in the train's frame, then since both light signals should have a velocity of c in the train's frame, you should predict both light signals reach the center at the same moment.

They will reach the center between the two at the same time, but since the train is moving toward one it will see the flash of that one first since the spped of light is finite.

 

[neutrino]

The effect would be reciprocal in that from the ships frame a will see the Earth's clock running the same rate as b sees but a will see the Earth clock lagging because it's further from the observer.

Earth clock lagging with respect to what? All the astronauts would observe is that the rate of the Earth's clock is slowed with respect to their own, not because of it's distance from them, but because it is moving with respect to them.

 

[JesseM]

Thanks JesseM, you're right. I think I've got it.

Ship a and ship b's clocks run at the same rate wrt to the observer but a lags behind b because it's further from the observer. That makes sense.

Not because it's further from the observer, no. This has nothing to do with the time the Earth observer sees the clocks reading, which is affected by light delays, it's about the coordinates the Earth-observer assigns to different ticks of each clock in his coordinate system after factoring out light delays (for example, if in 2007 I see an event 7 light-years away through my telescope, I assign that event a time-coordinate of 2000). Also, it will always be the trailing clock whose time is ahead in the Earth observer's frame, even if the trailing clock is further away from the Earth because the two clocks are moving towards Earth rather than away from it.

The effect would be reciprocal in that from the ships frame a will see the Earth's clock running the same rate as b sees but a will see the Earth clock lagging because it's further from the observer.

Since the ships are at rest relative to one another, each would measure the other ship's clock to be synchronized with their own clock and ticking at the same rate, in terms of the time-coordinates they assign to each tick in their frame after factoring out light delays. It is reciprocal in the sense that if you had two clocks at rest in the Earth's frame and synchronized in that frame, in the frame of a or b those two clocks would be out-of-sync. And it's probably best not to think of this as a natural "effect" in the same sense as time dilation, it's a consequence of the synchronization convention each observer chooses...for example, while it's true that if you accelerate a clock to a new speed it will "naturally" tick at the rate predicted by the time dilation equation for that new speed, it is not true that if you accelerate two clocks to a new speed they'll "naturally" remain synchronized in their own rest frame (it would depend on the specific details of how each one was accelerated), you'd have to resynchronize them manually.

 

[JesseM]

They will reach the center between the two at the same time, but since the train is moving toward one it will see the flash of that one first since the spped of light is finite.

But the whole point is to think how this looks in the train's rest frame, not just the frame where the train is moving. In the train's rest frame, the train is not moving toward either flash, it's at rest, and the light from each flash must travel at the same speed in the train's frame too. The only way to understand why the light from each flash doesn't reach the center of the train at the same time is to point to the relativity of simultaneity, which means that if the two flashes happened at the same time in the frame where the train is moving, they must have happened at different times in the train's rest frame. There is nothing like this in Newtonian mechanics, which would normally be used to analyze a situation like the one where two baseballs are thrown at a person running between them.

 

[paw]

Not because it's further from the observer, no. This has nothing to do with the time the Earth observer sees the clocks reading, which is affected by light delays, it's about the coordinates the Earth-observer assigns to different ticks of each clock in his coordinate system after factoring out light delays (for example, if in 2007 I see an event 7 light-years away through my telescope, I assign that event a time-coordinate of 2000).

Ok, I can understand there are conventions, but isn't it the same as saying the light delay is responsible? I mean if ship a is ahead of ship b by one light second and the ships clocks are syncronized then an Earth observer who saw b's clock as reading 12:00:01 would see a's clock as reading 12:00:00. Am I right?

From the ships frame, if b saw Earth's clock reading 12:00:01 then a would se Earth's clock reading 12:00:00. Is this right?

 

[JesseM]

Ok, I can understand there are conventions, but isn't it the same as saying the light delay is responsible? I mean if ship a is ahead of ship b by one light second and the ships clocks are syncronized then an Earth observer who saw b's clock as reading 12:00:01 would see a's clock as reading 12:00:00. Am I right?

No, if that were true, then that would be equivalent to saying the clocks are synchronized in the Earth's frame. That's what it means to be synchronized in a given frame--that once you factor out the light delays, assuming all light moves at c in your frame, both clocks showed the same reading at the same time-coordinate in your frame.

Suppose a and b are 10 light seconds apart in their rest frame, and their clocks are synchronized in their frame. Suppose they're moving at 0.6c in the Earth's frame. This means that, because of length contraction, in the Earth's frame they'll be measured to be just 8 light seconds apart. Does this mean that if the Earth observer looks through his telescope, he'll see the farther clock 8 seconds behind the closer clock? No, because this would mean that the clocks would be synchronized in his frame. In fact, if the two ships are moving towards him, so the closer ships is in the lead, when he looks through his telescope he'll see the farther clock as only 2 seconds behind the closer clock; so when he factors out the fact that it took light from the farther clock an extra 8 seconds to reach him, he'll conclude the farther clock must "really" be 6 seconds ahead of the closer clock in his frame. On the other had, if the two ship are moving away from him, so the farther ship is in the lead, then when he looks through his telescope he'll see the farther closer as a full 14 seconds behind the closer clock; when he factors out the light-delay, he'll conclude that the farther clock is "really" only 6 seconds ahead of the closer clock in his frame. In both cases, what he finds is that the clock that's in the lead has a time that's behind the the clock in the rear, by an amount of vx/c^2 = (0.6c)(10 light seconds)/c^2 = 6 seconds.

 

[paw]

Arrrggggg, it's so easy to confuse intuition with facts. Thanks, I think I have the full picture now. I should remember, 'when in doubt, calculate'.

 

[Santural]

...
Would someone explain paw's question and Jesse's answer? My thirst for knowledge drives me on!

 

[neutrino]

...
Would someone explain paw's question and Jesse's answer? My thirst for knowledge drives me on!

(Assuming that you're referring to post #28)

Paw wants to know whether synchronised clocks go out-of-synch for an observer in relative motion due to the time it takes for light to travel from the clocks, because they are placed at different distances to the observer.

The short answer is No. As for an explanation, I don't think I can do it better than JesseM.

 

[Santural]

And so Santural understood the question.
And so he understood the answer.
And so he was happy again.